What is (\partial)/(\partial y)((z)(tx^2+yz-2)) ?

The answer to (\partial)/(\partial y)((z)(tx^2+yz-2)) is z^2

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(\partial)/(\partial y)((z)(tx^2+yz-2))

\frac{\partial}{\partial y} ((z) (t x^{2} + yz - 2))

z^{2}

Solution steps

\frac{\partial}{\partial y} ((z) (t x^{2} + yz - 2))

Treat

z, t, x

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= z \frac{\partial}{\partial y} (t x^{2} + yz - 2)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= z (\frac{\partial}{\partial y} (t x^{2}) + \frac{\partial}{\partial y} (yz) - \frac{\partial}{\partial y} (2))

\frac{\partial}{\partial y} (t x^{2}) = 0

\frac{\partial}{\partial y} (yz) = z

\frac{\partial}{\partial y} (2) = 0

= z (0 + z - 0)

Simplify

z (0 + z - 0) : z^{2}

= z^{2}

\frac{d}{d t} ((x) (4 t + 3))

(e^{x})^{2}

\frac{( 20 x ^{3} - 30 x ^{2} + 12 x - 1 ) ^{2}}{400}

- \frac{x ^{2} ( 2 x ^{6} + 7 )}{196}

(1 + csc (y))^{2}

What is (\partial)/(\partial y)((z)(tx^2+yz-2)) ?
The answer to (\partial)/(\partial y)((z)(tx^2+yz-2)) is z^2